Question

For the reaction
$H_2+I_2\rightleftharpoons 2HI$
The value of equilibrium constant is $\mathrm{9.0.}$ The degree of dissociation of $HI$ will be

• A. 2
• B. 2/5
• C. 5/2
• D. 1/2

Answer 1

Answer: (B)

Equilibrium constant of the reaction

$H_2+I_2\rightleftharpoons 2HI\text{ is }9.0$

So the equilibrium constant for the dissociation of $HI$, i.e.

$2HI\rightleftharpoons H_2+I_2$ will be $1/9$



$K_C=\frac{x}{2}=\frac{x}{2}\frac{1}{(1-x)}\times \frac{1}{(1-x)}$

$\frac{1}{9}=\frac{x^2}{2\times 2(1-x{)}^2};\frac{1}{3}=\frac{x}{2(1-x)}$

$\Rightarrow x=2/5$